Q.

If the rate constant k for the decomposition of hydrocarbon related with temperature is 1011⋅e−28000T, then the activation energy (Ea) is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

-28000 R

b

28000R

c

28000 x R

d

-28000R

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Arrhenius equation:k=Ae−EaRTk=1011⋅e−28000/T∴ −EaRT=−28000TEa = 28000 × R.
Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon